Amongst `TiF_(6)^(2-), CoF_(6)^(3-), Cu_(2)Cl_(2) and NiCl_(4)^(2-)` the colourless species are:

Options

A. `CoF_(6)^(3-) and NiCl_(4)^(2-)`
B. `TiF_(6)^(2-) and CoF_(6)^(3-)`
C. `Cu_(2)Cl_(2) and NiCl_(4)^(2-)`
D. `TiF_(6)^(2-) and Cu_(2)Cl_(2)`

Answer

To determine which of the given species are colorless, we need to analyze the electronic configurations of the transition metals in each compound and check for the presence of unpaired electrons. The presence of unpaired electrons can lead to d-d transitions, which are responsible for the color in transition metal complexes.

### Step-by-Step Solution:

1. **Identify the oxidation states and electronic configurations:**
- **TiF₆²⁻:** Titanium is in the +4 oxidation state. The electronic configuration of titanium (Ti) is [Ar] 3d² 4s². In the +4 state, it loses all four valence electrons, resulting in 3d⁰.
- **CoF₆³⁻:** Cobalt is in the +3 oxidation state. The electronic configuration of cobalt (Co) is [Ar] 3d⁷ 4s². In the +3 state, it loses two 4s electrons and one 3d electron, resulting in 3d⁶.
- **Cu₂Cl₂:** Copper is in the +1 oxidation state (as Cu₂ means two Cu atoms, each in +1 state). The electronic configuration of copper (Cu) is [Ar] 3d¹⁰ 4s¹. In the +1 state, it loses one 4s electron, resulting in 3d¹⁰.
- **NiCl₄²⁻:** Nickel is in the +2 oxidation state. The electronic configuration of nickel (Ni) is [Ar] 3d⁸ 4s². In the +2 state, it loses two 4s electrons, resulting in 3d⁸.

2. **Determine the presence of unpaired electrons:**
- **TiF₆²⁻ (3d⁰):** No unpaired electrons (colorless).
- **CoF₆³⁻ (3d⁶):** Has unpaired electrons (colorful due to d-d transitions).
- **Cu₂Cl₂ (3d¹⁰):** No unpaired electrons (colorless).
- **NiCl₄²⁻ (3d⁸):** Has unpaired electrons (colorful due to d-d transitions).

3. **Conclusion:**
The colorless species among the given options are **TiF₆²⁻** and **Cu₂Cl₂**.

### Final Answer:
The colorless species are: **TiF₆²⁻ and Cu₂Cl₂.**

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